N cars are going to the same destination along a one lane road. The destination is target miles away.
Each car i has a constant speed speed[i] (in miles per hour), and initial position position[i] miles towards the target along the road.
A car can never pass another car ahead of it, but it can catch up to it, and drive bumper to bumper at the same speed.
The distance between these two cars is ignored - they are assumed to have the same position.
A car fleet is some non-empty set of cars driving at the same position and same speed. Note that a single car is also a car fleet.
If a car catches up to a car fleet right at the destination point, it will still be considered as one car fleet.
How many car fleets will arrive at the destination?
Example 1:
Input: target = 12, position = [10,8,0,5,3], speed = [2,4,1,1,3] Output: 3 Explanation: The cars starting at 10 and 8 become a fleet, meeting each other at 12. The car starting at 0 doesn‘t catch up to any other car, so it is a fleet by itself. The cars starting at 5 and 3 become a fleet, meeting each other at 6. Note that no other cars meet these fleets before the destination, so the answer is 3.
Note:
0 <= N <= 10 ^ 40 < target <= 10 ^ 60 < speed[i] <= 10 ^ 60 <= position[i] < target- All initial positions are different.
N辆车沿着一条车道驶向位于target英里之外的共同目的地。每辆车i以恒定的速度speed[i]（英里/小时），从初始位置 position[i]（英里）沿车道驶向目的地。
一辆车永远不会超过前面的另一辆车，但它可以追上去，并与前车以相同的速度紧接着行驶。此时，我们会忽略这两辆车之间的距离，也就是说，它们被假定处于相同的位置。车队是一些由行驶在相同位置、具有相同速度的车组成的非空集合。注意，一辆车也可以是一个车队。即便一辆车在目的地才赶上了一个车队，它们仍然会被视作是同一个车队。求会有多少车队到达目的地？
解法：先把车按照位置进行排序，然后计算出每个车在无阻拦的情况下到达终点的时间，如果后面的车到达终点所用的时间比前面车小，那么说明后车会比前面的车先到，由于后车不能超过前车，所以这种情况下就会合并成一个车队。用栈来存，对时间进行遍历，对于那些应该合并的车不进栈就行了，最后返回栈的长度。或者直接用一个变量存最近前车到达时间，用另一变量记录车队的数量，如果循环的时间大于记录的前车时间，则当前的车不会比之前的车先到达，为一个新车队，更新变量。
Java:
public int carFleet(int target, int[] pos, int[] speed) {
TreeMap<Integer, Double> m = new TreeMap<>();
for (int i = 0; i < pos.length; ++i) m.put(-pos[i], (double)(target - pos[i]) / speed[i]);
int res = 0; double cur = 0;
for (double time : m.values()) {
if (time > cur) {
cur = time;
res++;
}
}
return res;
}
Java:
class Solution {
public int carFleet(int target, int[] position, int[] speed) {
int N = position.length;
int res = 0;
//建立位置-时间的N行2列的二维数组
double[][] cars = new double[N][2];
for (int i = 0; i < N; i++) {
cars[i] = new double[]{position[i], (double) (target - position[i]) / speed[i]};
}
Arrays.sort(cars, (a, b) -> Double.compare(a[0], b[0]));
double cur = 0;
//从后往前比较
for (int i = N - 1; i >= 0; i--) {
if (cars[i][1] > cur) {
cur = cars[i][1];
res++;
}
}
return res;
}
}
Python:
def carFleet(self, target, pos, speed):
time = [float(target - p) / s for p, s in sorted(zip(pos, speed))]
res = cur = 0
for t in time[::-1]:
if t > cur:
res += 1
cur = t
return res
Python:
class Solution:
def carFleet(self, target, position, speed):
"""
:type target: int
:type position: List[int]
:type speed: List[int]
:rtype: int
"""
cars = [(pos, spe) for pos, spe in zip(position, speed)]
sorted_cars = sorted(cars, reverse=True)
times = [(target - pos) / spe for pos, spe in sorted_cars]
stack = []
for time in times:
if not stack:
stack.append(time)
else:
if time > stack[-1]:
stack.append(time)
return len(stack)
Python:
# Time: O(nlogn)
# Space: O(n)
class Solution(object):
def carFleet(self, target, position, speed):
"""
:type target: int
:type position: List[int]
:type speed: List[int]
:rtype: int
"""
times = [float(target-p)/s for p, s in sorted(zip(position, speed))]
result, curr = 0, 0
for t in reversed(times):
if t > curr:
result += 1
curr = t
return result
C++:
int carFleet(int target, vector<int>& pos, vector<int>& speed) {
map<int, double> m;
for (int i = 0; i < pos.size(); i++) m[-pos[i]] = (double)(target - pos[i]) / speed[i];
int res = 0; double cur = 0;
for (auto it : m) if (it.second > cur) cur = it.second, res++;
return res;
}
原文地址：https://www.cnblogs.com/lightwindy/p/9795809.html